MOOCULUS微积分-2: 数列与级数学习笔记 6. Power series

此课程(MOOCULUS-2 "Sequences and Series")由Ohio State University于2014年在Coursera平台讲授。

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本系列学习笔记PDF下载（Academia.edu） MOOCULUS-2 Solution

Summary

• Let $(a_n)$ be a sequence of real numbers starting with $a_0$. Then the power series associated to $(a_n)$ is $$\sum_{n=0}^\infty a_n \, x^n.$$ Note that $a_n$ does not depend on $x$.
• The set of values of $x$ for which the series $$\sum_{n=0}^\infty a_n \, x^n$$ converges is the interval of convergence. That is, by ratio test we have $$\lim_{n\to\infty}{|a_{n+1}\cdot x^{n+1}|\over|a_n\cdot x^{n}|}=|x|\cdot\lim_{n\to\infty}{|a_{n+1}|\over|a_n|} < 1$$ it will converge. Technically, $${1\over R}=\lim_{n\to\infty}{|a_{n+1}|\over |a_n|}$$
• For a power series, the interval of convergence is, in fact, an interval. It has the form $(-R,R)$ or $[-R,R)$ or $(-R,R]$ or $[-R,R]$. In short, it is centered around $0$.
• In the interval of convergence of a power series, the value $R$ is called the radius of convergence of the series.
• Let $(a_n)$ be a sequence of real numbers starting with $a_0$. Then the power series centered at $c$ and associated to $(a_n)$ is the series $$\sum_{n=0}^\infty a_n \, (x-c)^n.$$ That is, the interval of convergence is $I=(c-R, c+R)$ (or include the endpoints).
• Suppose the power series $$f(x)=\sum_{n=0}^\infty a_n(x-a)^n=a_0+a_1\cdot(x-a)+a_2\cdot(x-a)^2+\cdots$$ has radius of convergence $R$. Then $$f'(x)=a_1+2a_2\cdot(x-a)+\cdots=\sum_{n=1}^\infty na_n(x-a)^{n-1}$$ $$\int f(x)\,dx = C+\sum_{n=0}^\infty {a_n\over n+1}(x-a)^{n+1}$$ for interval of $x$ in the interval $(a-R, a+R)$. These two new series have radius of convergence $R$, just like the original series.

Exercises 6.3

Find the radius and interval of convergence for each series. In exercises 3 and 4, do not attempt to determine whether the endpoints are in the interval of convergence.

1. $$\sum_{n=0}^\infty n x^n$$  

Solution: $${1\over R}=\lim_{n\to\infty}{|a_{n+1}|\over|a_n|}=\lim_{n\to\infty}{n+1\over n}=1$$ Thus $R=1$. When $x=\pm1$, the series are $\sum_{n=0}^{\infty}n$ and $\sum_{n=0}^{\infty}(-1)^n\cdot n$, which are diverge. Therefore the interval of convergence is $I=(-1, 1)$.

2. $$\sum_{n=0}^\infty {x^n\over n!}$$  

Solution: $${1\over R}=\lim_{n\to\infty}{|a_{n+1}|\over|a_n|}=\lim_{n\to\infty}{n!\over (n+1)!}=\lim_{n\to\infty}{1\over n+1}=0$$ Thus $R=\infty$ and the interval of convergence is $I=(-\infty, \infty)$.

3. $$\sum_{n=1}^\infty {n!\over n^n}x^n$$  

Solution: $${1\over R}=\lim_{n\to\infty}{|a_{n+1}|\over|a_n|}=\lim_{n\to\infty}{(n+1)!\over(n+1)^{n+1}}\cdot{n^n\over n!}=\lim_{n\to\infty}({n\over n+1})^n={1\over e}$$ Thus $R=e$ and the interval of convergence is $I=(-e, e)$.

4. $$\sum_{n=1}^\infty {n!\over n^n}(x-2)^n$$  

Solution: $${1\over R}=\lim_{n\to\infty}{|a_{n+1}|\over|a_n|}=\lim_{n\to\infty}{(n+1)!\over(n+1)^{n+1}}\cdot{n^n\over n!}=\lim_{n\to\infty}({n\over n+1})^n={1\over e}$$ Thus $R=e$ and the interval of convergence is $I=(2-e, 2+e)$.

5. $$\sum_{n=1}^\infty {(n!)^2\over n^n}(x-2)^n$$  

Solution: $${1\over R}=\lim_{n\to\infty}{|a_{n+1}|\over|a_n|}=\lim_{n\to\infty}{[(n+1)!]^2\over(n+1)^{n+1}}\cdot{n^n\over(n!)^2}=\lim_{n\to\infty}{(n+1)^2\over(n+1)^{n+1}}\cdot n^n$$ $$=\lim_{n\to\infty}(n+1)\cdot({n\over n+1})^n={1\over e}\cdot\lim_{n\to\infty}(n+1)=\infty$$ Thus $R=0$ and it converges only on $x=2$ and diverges otherwise.

6. $$\sum_{n=1}^\infty {(x+5)^n\over n(n+1)}$$

Solution: $${1\over R}=\lim_{n\to\infty}{|a_{n+1}|\over|a_n|}=\lim_{n\to\infty}{n(n+1)\over(n+1)(n+2)}=1$$ Thus $R=1$ and the endpoints are $x_1=-5-1=-6$ and $x_2=-5+1=-4$. Both of them are convergent. So the interval of convergence is $I=[-6, -4]$.

7. Find a power series with radius of convergence $0$.  

Solution:

There are many choices---for instance, see Exercise 5---alternatively $\sum_{n=0}^\infty n! \cdot x^n$ also works.

Exercises 6.4

1. Find a series representation for $\log 2$.

Solution:

Begin with the geometric series, namely $${1\over1-x}=\sum_{n=0}^{\infty}x^n\Rightarrow \int {1\over1-x}dx=-\log|1-x|= \sum_{n=0}^{\infty}{1\over n+1} x^{n+1}$$ So $x=-1$ and the result is $$\log2=-\sum_{n=0}^{\infty}{(-1)^{n+1}\over n+1} = \sum_{n=0}^{\infty}{(-1)^{n} \over n+1}$$

2. Find a power series representation for $1/(1-x)^2$.

Solution: $${1\over1-x}=\sum_{n=0}^{\infty}x^n\Rightarrow ({1\over1-x})'={1\over(1-x)^2}=\sum_{n=1}^{\infty}nx^{n-1}$$

3. Find a power series representation for $2/(1-x)^3$.

Solution: $${1\over1-x}=\sum_{n=0}^{\infty}x^n\Rightarrow ({1\over1-x})'={1\over(1-x)^2}=\sum_{n=1}^{\infty}nx^{n-1}$$ $$\Rightarrow ({1\over1-x})''={2\over(1-x)^3}=\sum_{n=2}^{\infty}n(n-1)x^{n-2}$$

4. Find a power series representation for $1/(1-x)^3$. What is the radius of convergence?

Solution:

According to the above exercise, we have $${1\over(1-x)^3}=\sum_{n=2}^{\infty}{n(n-1)\over2}x^{n-2}=\sum_{n=0}^{\infty}{(n+1)(n+2)\over2}x^{n}$$ And $${1\over R}=\lim_{n\to\infty}{|a_{n+1}|\over|a_n|}=\lim_{n\to\infty}{(n+2)(n+3)\over(n+1)(n+2)}=1$$ Thus the radius is $R=1$.

5. Find a power series representation for $\int\log(1-x)\,dx$.

Solution: $$\log(1-x)=-\int {1\over1-x}dx=-\int\sum_{n=0}^{\infty}x^n dx=\sum_{n=0}^{\infty}{-1\over n+1}x^{n+1}$$ $$\Rightarrow \int\log(1-x)dx=\int\sum_{n=0}^{\infty}{-1\over n+1}x^{n+1}dx=C+\sum_{n=0}^{\infty}{-1\over(n+1)(n+2)}x^{n+2}$$

Additional Exercises

1. For which real number $x$ does the series $$\sum_{m=4}^{\infty}{({1\over6})^m\cdot x^m\over7m}$$ converge.

Solution:

Let $a_m={({1\over6})^m\over7m}$, we have $${1\over R}=\lim_{m\to\infty}{|a_{m+1}|\over |a_m|}=\lim_{m\to\infty}{({1\over6})^{m+1}\over7(m+1)}\cdot{7m\over({1\over6})^m}={1\over6}$$ Thus $R=6$. The endpoints are $x_1=-6$ and $x_2=6$. When $x=-6$, we have $$\sum_{m=4}^{\infty}{({1\over6})^m\cdot x^m\over7m}=\sum_{m=4}^{\infty}{(-1)^m\over7m}$$ which is an alternating harmonic series, and it converges. When $x=6$, we have $$\sum_{m=4}^{\infty}{({1\over6})^m\cdot x^m\over7m}=\sum_{m=4}^{\infty}{1\over7m}$$ which is harmonic series, and it diverges. Thus the interval of converges is $I=[-6, 6)$.

2. Which is the radius of convergence of the series $$\sum_{n=4}^{\infty}{(8^n+n)\cdot x^n\over3n}$$

Solution:

Let $a_n={8^n+n\over3n}$, we have $${1\over R}=\lim_{n\to\infty}{|a_{n+1}|\over|a_n|}=\lim_{n\to\infty}{8^{n+1}+n+1\over3(n+1)}\cdot{3n\over 8^n+n}=8$$ Thus the radius is $R={1\over8}$.

3. $$f(x)=\sum_{n=0}^{\infty}-{(5n+3)x^n\over2n-3}$$ Consider $f'(x)$.

Solution: $$f'(x)=\sum_{n=1}^{\infty}-{5n+3\over2n-3}\cdot n\cdot x^{n-1}=\sum_{n=0}^{\infty}-{5(n+1)+3\over2(n+1)-3}\cdot (n+1)\cdot x^{n+1-1}$$ $$=\sum_{n=0}^{\infty}-{5n+8\over2n-1}\cdot (n+1)\cdot x^{n}$$

4. Suppose $$\sum_{n=1}^{\infty}b_n={2\over(2x-1)^2}$$ Find an expression of $b_n$ (involve $x$).

Solution:

Let $F(x)=\sum_{n=1}^{\infty}b_n={2\over(2x-1)^2}$, we have $$\int F(x)=\int {2\over(2x-1)^2} dx=\int{d(2x-1)\over(2x-1)^2}={1\over1-2x}=f(x)$$ On the other hand, $$f(x)={1\over1-2x}=\sum_{n=0}^{\infty}(2x)^n$$ Thus $$F(x)=f'(x)=\sum_{n=1}^{\infty}2^n\cdot n\cdot x^{n-1}$$ $$\Rightarrow b_n=2^n\cdot n\cdot x^{n-1}$$

5. Suppose $$\sum_{n=1}^{\infty}b_n={9x\over(9x^2-1)^2}$$ Find an expression of $b_n$ (involve $x$).

Solution:

Let $F(x)=\sum_{n=1}^{\infty}b_n={9x\over(9x^2-1)^2}$, we have $$\int F(x)=\int {9x\over(9x^2-1)^2}dx={1\over2}\cdot\int {d(9x^2-1)\over(9x^2-1)^2}={1\over2}\cdot{1\over1-9x^2}=f(x)$$ On the other hand, $$f(x)={1\over2}\cdot\sum_{n=0}^{\infty}(9x^2)^n$$ Thus $$F(x)=f'(x)={1\over2}\cdot\sum_{n=1}^{\infty}9^n\cdot 2n\cdot x^{2n-1}$$ $$\Rightarrow b_n=9^n\cdot n\cdot x^{2n-1}$$

6. Consider the function $$f(t)=\int_{0}^{t}e^{-x^2}dx$$ Compute $f({3\over2})$ to within ${1\over2}$.

Solution:

Note that the power series (Taylor series) of $e^x$ is $$e^x=\sum_{n=0}^{\infty}{x^n\over n!}$$ Thus we have $$e^{-x^2}=\sum_{n=0}^{\infty}{{(-x^2)}^n\over n!}=\sum_{n=0}^{\infty}{(-1)^n\cdot x^{2n}\over n!}$$ $$\Rightarrow f(t)=\int_{0}^{t}e^{-x^2}dx=\int_{0}^{t}\sum_{n=0}^{\infty}{(-1)^n\cdot x^{2n}\over n!}dx$$ $$=\sum_{n=0}^{\infty}{(-1)^n\over n!}\cdot\int_{0}^{t}x^{2n}dx=\sum_{n=0}^{\infty}{(-1)^n\over n!}\cdot{1\over 2n+1}x^{2n+1}\Big|_{0}^{t}$$ $$=\sum_{n=0}^{\infty}{(-1)^n\over n!}\cdot{1\over 2n+1}t^{2n+1}$$ $$\Rightarrow f({3\over2})=\sum_{n=0}^{\infty}{(-1)^n\over n!}\cdot{1\over 2n+1}({3\over2})^{2n+1}=a_n$$ Our aim is to find an $|a_n| < 0.5$ and by computing in R:

f = function(x) (-1)^x / factorial(x) * 1 / (2 * x + 1) * (3/2)^(2 * x + 1)

for (i in 0:100) {

  if (f(i) < 0.5 & f(i) > -0.5) {

    print(i)

    print(f(x = 0:i))

    print((sum(f(0:i)) + sum(f(0:(i-1)))) / 2) 

    break

  }

}

# [1] 3

# [1]  1.500000 -1.125000  0.759375 -0.406808

# [1] 0.930971

That is, $$a_0=1.5,\ a_1=-1.125,\ a_2=0.759375,\ a_3=-0.406808 \in (-0.5, 0.5)$$ Thus the value within 0.5 is $${1\over2}\cdot (s_2+s_3)=0.930971$$